Why should you do you should

lozanna [386]

Answer:

D) a Battery

Explanation:

The best real-life example of direct current is a battery. Batteries have positive and negative terminals on a battery, the electrons in the wires will begin to flow to produce a current.

6 0

3 years ago

An object weighs 10N on earth .what is the objects weight on a planet one tenth the earths mass and one half its radius?

earnstyle [38]

We know, weight = mass * gravity
10 = m * 9.8
m = 10/9.8 = 1.02 Kg

Now, Let, the gravity of that planet = g'
g' = m/r² [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)² [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g
g' = 2/5 * 9.8
g' = 3.92

Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N

In short, Your Answer would be 4 Newtons

Hope this helps!

3 0

3 years ago

A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr

Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

brainly.com/question/15365772

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7 0

2 years ago

A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from

s2008m [1.1K]

E = kQ 

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = 9e9 * 300e^-6 

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0

2 years ago

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re

valentinak56 [21]

Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

5 0

3 years ago